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at $9, etc. Thus, as the number of yards is changed, the price paid changes also. Either depends on the other, i.e., to a given number of yards corresponds a certain price and to a given price corresponds a certain number of yards. Moreover, the dependence is such that by doubling the number of yards, the price is doubled. If the number of yards is trebled, the price is trebled, etc. The number of yards and the price paid are said to be directly proportional to each other. One is said to vary directly as the other.

This type of variation is very common. For example, the more workmen a factory employs, the greater is the pay roll; the time a train travels at a uniform rate varies as the distance; a man's pay depends on the number of days he works. Examples of direct variations are to be studied in this chapter.

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FORMULA FOR UNIFORM MOTION

142. How to represent the formula graphically. We have seen that the formula for uniform motion is d=rt, where d is the number of units (inches, feet, meters) of distance, r the rate, and t the number of units (seconds, hours, days) of time. If a train travels at the rate of 30 miles an hour, the number of miles is 30 times the number of hours, i.e., d=30t. As t varies, d also varies; d but the ratio is constant and equal to 30. Hence we

t may say that d varies directly as t ($141).

The relation d=30t may be represented graphically as follows: First make a table of several pairs of corresponding values of d and t (Fig. 265). Thus, let t=1, then d=30; if t=2, then d=60; etc.

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The next step is to draw the two reference axes, OT and OD.

Convenient units are now chosen for representing graphically the number pairs in the table. These units are marked off on the axes and the points corresponding to number pairs are marked down, or plotted. Thus, to plot the pair (1, 30) pass from 0 one unit to the right, and then up 30 units. To plot (2, 60) pass from 0 two units to the right and then up 60 units.

Finally, the line passing through the points is drawn. This is the required graph.

The graph may be used to find values of d for given values of t; or to find t when d is given. Thus, when t=21, pass from 0 two and one-half units to the right to E. From E pass upward to F. This vertical height may now be read off on the line OD. The result is d=75.

To find t when d=85, pass from O upward along the line OD a distance of 85 units to G. Then pass to the right to the graph to H, and from H pass downward to the line OT. The result is t=2.8.

EXERCISES

1. Make a graph of the equation d=20t, following the directions given above.

2. From the graph of Exercise 1, find the values of d corresponding to the values t=1), 31, 21; find t when d=40, 64, 98.

3. Make a graph of the equation d=18.5t.

Using the formula d=rt, i.e., “Distance is equal to rate times the time," solve the following problems:

4. The speed of a train is 28 miles an hour. If the train leaves the station at 2 P.M., how far from the station will it be at 3:00 P.M.; 3:30 P.M.; 4:20 P.M.; 5:00 P.M.?

5. Sound travels about 1080 feet per second. How far away is a stroke of lightning if the thunder clap is heard 30 seconds after the flash; 20 seconds after the flash?

6. From the news- SETHKIEWICZ LOPES TO TRI

UMPH IN RACE OVER 15 paper clipping at the

MILE COURSE right, find out how fast

John R. Sethkiewicz, Illinois A. C., won the winner could run.

the fifteen mile distance run of the Michi

gan-Irving C. C. yesterday. The victor cov(1:16:30 means “one

ered the course in 1:16:30, finishing, well ahead of Frank Stehlak of Palmer Park,

who was second. hour, sixteen minutes, thirty seconds.”)

McHALE WINS AUTO RACE

Kalamazoo, Mich., Aug. 6.- Special.}7. From the news

Bud E. McHale of Detroit, driving a Ford

special, won the 100-mile auto race here this item at the right, de

afternoon. Benny Lawell of Columbus, O.,

finished second. Time 1:33:31. termine how many miles per hour Lawell of Columbus was traveling in his car.

PERCENTAGE FORMULA

143. Rate, or per cent. In a written test a pupil made a score of 22. The total possible score was 25. How may his achievement be measured?

Solution: As a measure of the pupil's achievement of 22 scores in a possible score of 25, we may use the ratio 23 Dividing 22 by 25, we have =.88 = 1948

100 This means that a score of 22 out of a possible score of 25 is equivalent to a score of 88 out of a possible 100. The number of "hundredths,” which in this case is 88, is the “grade” of the pupil in the test.

Grades are often expressed as "hundredths,” because it is then easy to compare the pupil's achievements in several tests. For example, if the same pupil in the next test has a score of 18 out of 20, his achievement 18 is equal to 10% when expressed in hundredths. This is higher than the 8.8 in the previous test.

The number of hundredths is the number of per cents ($19). Thus we say that the pupil's achievements of 22 scores out of 25, or of 18 out of 20, are respectively 88 per cent and 90 per cent. The symbol for per cent is % ($19). With this symbol "88 per cent” is written

88%

The number of per cent, 88, is called rate; ** is the rate per cent.

EXERCISES

1. In the show window of a furniture company the following electric sign is displayed:

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2. A clothing firm advertises for a special sale a reduction of 40% on all suits and overcoats. What will be the reduction on a suit which is usually sold for $38?

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